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Physics 9 Fall 2009
Midterm 1 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck!
- Recalling that the electric field at a distance r inside a uniformly charged sphere, of radius R, carrying charge Q, is
E^ ~ (~r) = Q 4 π 0 R^3
rr,ˆ
determine the electric potential V (r) at a point r inside the sphere, relative to infinity. ————————————————————————————————————
Solution
The potential is given in terms of the electric field as
V (r) = −
∫ (^) r
∞
E^ ~ · d~`.
Inside the sphere, the electric field is given as above. Outside the sphere, the electric field is just that of a point charge,
E^ ~ (~r) = 1 4 π 0
Q
r^2
ˆr.
So, in order to find the potential, we have to integrate the electric field over both regions, breaking up the integral from ∞ to R, the radius of the sphere. Then we integrate the internal field from R to r. Since we’re integrating over a radial direction, we have ˆr · d~` = dr. So, integrating gives
V (r) = −
[
4 π 0
Q
∫ R
∞
4 π 0
dr r^2
4 π 0
Q
R^3
∫ (^) r
R
rdr
]
Performing the integrations gives
V (r) = −
[
1 4 π 0 Q^
∫ R
∞
1 4 π 0
dr r^2 +^
1 4 π 0
Q R^3
∫ (^) r R rdr
]
= (^4) πQ 01 r
∣R
∞ −^
1 2
Q 4 π 0 R^3 r
2 ∣∣r R = (^4) π^10 QR − (^124) π^10 R^ Q 3 (r^2 − R^2 ).
Combining the first and last terms gives the potential inside
V (r) =
Q
8 π 0 R
r^2 R^2
- Show that the electric field due to an infinitely long, uniformly charged thin cylindrical shell of radius a having a surface charge density QA = η is given by the following expressions: E =
0 for 0 ≤ r < a ηa 0 r for^ r^ ≥^ a. ————————————————————————————————————
Solution
We can solve this problem using Gauss’s law, which says that ∮ E^ ~ · d A~ = Qencl 0
The situation is as seen in the figure to the right. Because of the cylindrical symmetry we choose a cylinder of radius r and length L as our Gaussian surface. As we’ve seen many times, the electric field is constant on the Gaussian surface and points along the direction of the normal to the surface, and so
a
r r
η
E^ ~ · d A~ = EA,
where A = 2πrL is the surface area through which the electric field is fluxing (note that the ends of the cylinder don’t contribute to the flux since the surface is perpendicular to the electric field there). So, the net flux is 2πrLE. Now we just need to work out the enclosed charge. For a Gaussian surface inside the cylinder, the enclosed charge is zero, since all the charge lies on the surface. So, inside the cylinder, 2πrLE = Qencl 0 = 0, and so inside the cylinder, E = 0. Outside the cylinder, a Gaussian surface of length L encloses a certain amount of charge Qencl. This can be given in terms of the surface charge density, η, writing Qencl = ηAencl, where Aencl is the area enclosed by the Gaussian surface. The area of the charge is just the area of the shell, Aencl = 2πaL, since all the charge is located on the surface at r = a. So, Qencl = η (2πaL). Putting everything together gives
(2πrL) E =
η (2πaL) 0
⇒ E =
ηa 0 r
So, we find E =
0 for 0 ≤ r < a ηa 0 r for^ r^ ≥^ a, which is exactly right.
- Consider the RC circuit shown in the dia- gram. In your homework, you showed that if the switch is closed at time t = 0, then the charge on the capacitor at time t was given by the expression
Q = Qmax
1 − e−t/τ^
where the time constant, τ ≡ RC.
R
S
^ C
E
(a) In terms of τ , how long does the capacitor take to charge to half of its maximum value? (b) What is the potential across the capacitor, ∆VC , at a time t? (c) What happens to the potential across the capacitor, ∆VC at very late times? (d) Show that Qmax = CE. (e) What is the current, I(t), in the capacitor at a time t in terms of R, C, and E?
————————————————————————————————————
Solution
(a) At a time t 0 , the charge is half of the maximum charge, so Q(t 0 ) = Qmax 2 = Qmax
1 − e−t^0 /τ^
. So, canceling the common factor of Qmax and rearranging gives e−t^0 /τ^ = 12. Taking natural logarithms gives t 0 = ln (2) τ ≈ 0. 69 τ. (b) Since Q = CV , the potential is just
V =
Q
C
Qmax C
1 − e−t/τ^
(c) At very late times the capacitor must saturate to the full potential given by the battery, E. (d) Using our result from part (c), as t → ∞, the exponential term vanishes, and so V → E = Qmax C , and so Qmax = CE. (e) The current in the capacitor is just I = dQdt. Taking the derivative gives, I(t) = d dt
Qmax
1 − e−t/τ^
= Qmax τ e−t/τ^. But, since Qmax = CE, and τ = RC, we find
I(t) =
E
R
e−t/RC^.
Extra Credit Question!!
The following is worth 10 extra credit points!
The potential energy be- tween a pair of neutral atoms or molecules is very well-approximated by the Lennard-Jones Potential, given by the expression
P E(r) = 4
[(
σ r
(σ r
) 6 ]
where and σ are constants, and r is the distance be- tween the molecules. The potential energy is plotted in the figure to the right. The vertical axis is in units of , while the horizontal axis is in units of σ.
Molcular Bond Energy
**-
-**
0
1
2
3
4
5
6
7
8
0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.
Distance
Energy
(a) Why does the potential energy approach zero as the distance gets bigger? (b) At what separation distance, in terms of σ and , is the potential energy zero? (c) At approximately what distance is the system in equilibrium? What is the po- tential energy at that distance? (Express your answers in terms of σ and .) (d) How much energy would you need to add to the system at equilibrium in order to break the molecular bonds holding it together? Why? (e) How much energy is released in the breaking of those molecular bonds? Why?
Note - no calculation is needed to answer these problems!
————————————————————————————————————
Solution
(a) As the two molecules get further apart, the attractive force between them gets weaker and weaker. When the are very far apart, they hardly interact at all - they are basically free molecules. The potential energy of a free particle is zero, since potential energy depends on the interaction between multiple particles. (b) We can just read the value off from the graph. We see that the potential energy crosses the x axis when x = 1, which means that r = σ. We can see this from the equation, too: setting r = σ gives P E(σ) = 0.
